Question

Solve for $x\in R{\log }_x(2x-\frac{3}{4})>2$

Answer 1

Base must be positive and not equal to one i.e x>0 ($\ne 1$)

Argument must also be positive i.e. 2x-$\frac{3}{4}>0$

For the required question,

1. If 1>base>0, 2x-$\frac{3}{4}<x^2$

2. If base>1, 2x-$\frac{3}{4}>x^2$

Solving these conditions we get x$\in (\frac{3}{8},\frac{1}{2})\cup (1,\frac{3}{2})$